# A current of 0.25 A flows in the main circuit. Now the resistance R is disconnected and then connected across the 4 ohm resistance. The current in the curcuit now is

_{net}

_{= }2R/(2+R)+4+3, = (9R+4)/(2+R),

ACCORDING TO CONDITION I=0.25, V= 2VOLT

0.25= 2(2+R)/(9R+4) (Since I=V/R)

on solving for R,

R = 2 ohm

now as per question this resistor is connected across resistor 4 ohm...

net R= 2+3+4*2/(4+2)= 6.333ohm

V

_{given}=2volt

so I = V/R = 2/6.33

I=0.31A

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